Proof 11: Sphere and ellipsoid volume

A sphere can be cut into slices. Each slice is almost a disk, or a very flat cylinder, except for the curved edge. If we cut it into many, many – say, $n$ – slices, then each slice is fairly thin, and the curve might not matter much. Let’s call the height of each slice $\mathrm dh$. The sphere volume is of course the total volume of all these flat cylinders.

V_\text{sphere}&=\underbrace{V_1+V_2+V_3+\cdots+V_n}_{n \text{ terms}}\\
~&= \underbrace{A_1 \mathrm dh+A_2 \mathrm dh +A_3 \mathrm dh +\cdots+A_n \mathrm dh }_{n \text{ terms}}\qquad\leftarrow \text{ Cylinder volumes, } V=A_\text{base}h\\
~&= \underbrace{\pi r_1^2 \mathrm dh+ \pi r_2^2 \mathrm dh + \pi r_3^2 \mathrm dh +\cdots+ \pi r_n^2 \mathrm dh }_{n \text{ terms}}\qquad\leftarrow \text{ Circle areas, } A=\pi r^2\\
~&= \pi \underbrace{(r_1^2+ r_2^2 + r_3^2 +\cdots+ r_n^2 )}_{n \text{ terms}} \mathrm dh\\
~&=\pi\int_{-r}^r r_\text{slice}^2\, \mathrm dh\qquad\leftarrow\text{ Summed up with an integral from bottom } -r \text{ to top } r\\
~&=\pi\int_{-r}^r r^2-h^2\, \mathrm dh\qquad\leftarrow \text{ Pythagoras’ Theorem, } r_\text{slice}^2+h^2=r^2\Leftrightarrow r_\text{slice}^2=r^2- h^2 \\
~&=\pi\left(\int_{-r}^r r^2\, \mathrm dh+ \int_{-r}^r -h^2\, \mathrm dh\right) \\
~&=\pi\left(\left[r^2h\right]_{-r}^r + \left[ -\frac13h^3 \right ] _{-r}^r \right) \qquad\leftarrow \text{ Polynomium integration, } \int x^k\, \mathrm dx=\frac1k x^{k+1} \\
~&=\pi\left(r^2r-(- r^2r )+\left(-\frac13r^3 \right )- \left(-\frac13(-r)^3 \right ) \right) \\
~&=\pi\left(2r^3-\frac13r^3 -\frac13r^3 \right) \\
~&=\pi\left(2r^3-\frac23r^3 \right) \\
~&=\pi\left(\frac63r^3-\frac23r^3 \right) \\
~&=\pi\frac43r^3\quad_\blacksquare \\

Now, imagine a sphere with a radius of 1, $r=1$ (called a unit sphere). Its volume is

$$V_\text{unit sphere}=\frac43\pi 1^3=\frac43 \pi$$

As in Proof 5, we can imagine this unit sphere stretched in 3 dimensions to form a larger shape. You could double the size sideways, triple it upwards/downwards and half it inwards/outwards, so $\frac43\pi\cdot2 \cdot 3 \cdot ½$. Each stretch expands (or shrinks) the volume a certain number of times.

In general, let’s symbolise the stretches as $r_1$, $r_2$ and $r_3$. If they are not the same, then this new shape is an ellipsoid. The volume of such stretched – or ‘squeezed’ – shape is then:

$$V_\text{ellipsoid}=\frac43\pi r_1 r_2 r_3\qquad\blacksquare$$

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